WWL Chen - Fundamentals of Analysis (Chapter 1)

FUNDAMENTALS OF ANALYSIS

W W L CHEN

c© W W L Chen, 1983, 2008.

This chapter originates from material used by the author at Imperial College, University of London, between 1981 and 1990.

It is available free to all individuals, on the understanding that it is not to be used for financial gain,

and may be downloaded and/or photocopied, with or without permission from the author.

However, this document may not be kept on any information storage and retrieval system without permission

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Chapter 1

THE NUMBER SYSTEM

1.1. The Real Numbers

In this chapter, we shall make a detailed study of some of the important properties of the real numbers.Most readers will be familiar with some of these properties, or have at least used most of them, perhapssometimes unaware of their generality. Throughout, we denote the set of all real numbers by R, andwrite a ∈ R to indicate that a is a real number.

We shall take an axiomatic approach to the real numbers. In other words, we offer no proof of theseproperties, and simply treat and accept them as given.

The first collection of properties of R is generally known as the Field axioms. They enable us to studyarithmetic.

FIELD AXIOMS.(A1) For every a, b ∈ R, we have a+ b ∈ R.(A2) For every a, b, c ∈ R, we have a+ (b+ c) = (a+ b) + c.(A3) For every a ∈ R, we have a+ 0 = a.(A4) For every a ∈ R, there exists −a ∈ R such that a+ (−a) = 0.(A5) For every a, b ∈ R, we have a+ b = b+ a.(M1) For every a, b ∈ R, we have ab ∈ R.(M2) For every a, b, c ∈ R, we have a(bc) = (ab)c.(M3) For every a ∈ R, we have a1 = a.(M4) For every a ∈ R such that a 6= 0, there exists a−1 ∈ R such that aa−1 = 1.(M5) For every a, b ∈ R, we have ab = ba.

(D) For every a, b, c ∈ R, we have a(b+ c) = ab+ ac.

Chapter 1 : The Number System page 1 of 13

Fundamentals of Analysis c© W W L Chen, 1983, 2008

Remark. The properties (A1)–(A5) concern the operation addition, while the properties (M1)–(M5)concern the operation multiplication. In the terminology of group theory, we say that the set R formsan abelian group under addition, and that the set of all non-zero real numbers forms an abelian groupunder multiplication. We also say that the set R forms a field under addition and multiplication. Theproperty (D) is called the Distributive law.

The second collection of properties of R is generally known as the Order axioms. They enable us tostudy inequalities.

ORDER AXIOMS.(O1) For every a, b ∈ R, exactly one of a < b, a = b, a > b holds.(O2) For every a, b, c ∈ R satisfying a > b and b > c, we have a > c.(O3) For every a, b, c ∈ R satisfying a > b, we have a+ c > b+ c.(O4) For every a, b, c ∈ R satisfying a > b and c > 0, we have ac > bc.

Remark. Clearly the Order axioms as given do not appear to include many other properties of the realnumbers. However, these can be deduced from the Field axioms and Order axioms.

Example 1.1.1. Suppose that the real number a > 0. Then the real number −a < 0. To see this, notefirst that by Axiom (A4), there exists −a ∈ R such that a+ (−a) = 0. Hence

0 = a+ (−a) from above,> 0 + (−a) by Axiom (O3),= (−a) + 0 by Axiom (A5),= −a by Axiom (A3),

as required.

Example 1.1.2. For every a ∈ R, we have a0 = 0. To see this, note first that a0 ∈ R, in view of Axiom(M1). On the other hand, it follows from Axioms (A3) and (D) that a0 = a(0 + 0) = a0 +a0. Note nextthat −(a0) ∈ R and a0 + (−(a0)) = 0, in view of Axiom (A4). Hence

0 = a0 + (−(a0)) from above,= (a0 + a0) + (−(a0)) from above,= a0 + (a0 + (−(a0))) by Axiom (A2),= a0 + 0 by Axiom (A4),= a0 by Axiom (A3),

as required.

Example 1.1.3. Suppose that the real number a > 0. Then the real number a−1 > 0. To see this, notefirst that by Axiom (M4), there exists a−1 ∈ R such that aa−1 = 1. Suppose on the contrary that it isnot true that a−1 > 0. Then it follows from Axiom (O1) that a−1 = 0 or a−1 < 0. If a−1 = 0, then

1 = aa−1 by Axiom (M4),= a0= 0 by Example 1.1.2,

and soa = a1 by Axiom (M3),

= a0 from above,= 0 by Example 1.1.2,



a contradiction. If a−1 < 0, then

0 = a0 by Example 1.1.2,= 0a by Axiom (M5),> a−1a by Axiom (O4),= aa−1 by Axiom (M5),= 1 by Axiom (M4),

and so0 = a0 by Example 1.1.2,> a1 from above,= a by Axiom (M3),

again a contradiction.

Example 1.1.4. Suppose that the real numbers a > 0 and b > 0. Then the real number ab > 0. To seethis, note first that by Axiom (M1), we have ab ∈ R. Suppose on the contrary that it is not true thatab > 0. Then it follows from Axiom (O1) that ab = 0 or ab < 0. Since b > 0, it follows from Axiom(O1) that b 6= 0, from Axiom (M4) that b−1 ∈ R, and from Example 1.1.3 that b−1 > 0. If ab = 0, then

a = a1 by Axiom (M3),= a(bb−1) by Axiom (M4),= (ab)b−1 by Axiom (M2),= 0b−1

= b−10 by Axiom (M5),= 0 by Example 1.1.2,

a contradiction. If ab < 0, then

a = a1 by Axiom (M3),= a(bb−1) by Axiom (M4),= (ab)b−1 by Axiom (M2),< 0b−1 by Axiom (O4),= b−10 by Axiom (M5),= 0 by Example 1.1.2,

again a contradiction.

Example 1.1.5. Suppose that a, b ∈ R and 0 < a < b. Then b−1 < a−1. To see this, note first fromExample 1.1.3 that a−1 > 0 and b−1 > 0, and from Example 1.1.4 that b−1a−1 > 0. Hence

b−1 = b−11 by Axiom (M3),= b−1(aa−1) by Axiom (M4),= b−1(a−1a) by Axiom (M5),= (b−1a−1)a by Axiom (M2),< (b−1a−1)b by Axiom (O4),= (a−1b−1)b by Axiom (M5),= a−1(b−1b) by Axiom (M2),= a−1(bb−1) by Axiom (M5),= a−11 by Axiom (M4),= a−1 by Axiom (M3),

as required.



An important subset of the set R of all real numbers is the set of all natural numbers, given by

N = {1, 2, 3, . . .}.

However, this definition does not bring out some of the main properties of the set N in a natural way.The following more complicated definition is therefore sometimes preferred.

AXIOMS OF THE NATURAL NUMBERS.(N1) 1 ∈ N.(N2) If n ∈ N, then the number n+ 1, called the successor of n, also belongs to N.(N3) Every n ∈ N other than 1 is the successor of some number in N.

(WO) Every non-empty subset of N has a least element.

Remark. The condition (WO) is called the Well-ordering principle.

To explain the significance of each of these four axioms, note first that Axioms (N1) and (N2) to-gether imply that N contains 1, 2, 3, . . . . However, these two axioms alone are insufficient to excludefrom N numbers such as 5.5. Now, if N contained 5.5, then by Axiom (N3), N must also contain4.5, 3.5, 2.5, 1.5, 0.5,−0.5,−1.5,−2.5, . . . , and so would not have a least element. We therefore excludethis possibility by stipulating that N has a least element. This is achieved by Axiom (WO).

It can be shown that Axiom (WO) implies the Principle of induction. The following two forms of thePrinciple of induction are particularly useful. In fact, both are equivalent to Axiom (WO).

PRINCIPLE OF INDUCTION (WEAK FORM). Suppose that the statement p(.) satisfies thefollowing conditions:(PIW1) p(1) is true; and(PIW2) p(n+ 1) is true whenever p(n) is true.Then p(n) is true for every n ∈ N.

PRINCIPLE OF INDUCTION (STRONG FORM). Suppose that the statement p(.) satisfies thefollowing conditions:

(PIS1) p(1) is true; and(PIS2) p(n+ 1) is true whenever p(m) is true for all m ≤ n.

Then p(n) is true for every n ∈ N.

Proof of the equivalence of the Well-ordering principle and the two Principles ofinduction. Our first step is to show that Axiom (WO) is equivalent to the Principle of induction(strong form) (PIS).

((WO) ⇒ (PIS)) Suppose that the conclusion of (PIS) does not hold. Then the subset

S = {n ∈ N : p(n) is false}

of N is non-empty. By Axiom (WO), S has a least element, n0 say. If n0 = 1, then clearly (PIS1) doesnot hold. If n0 > 1, then p(m) is true for all m ≤ n0 − 1 but p(n0) is false, contradicting (PIS2).

((PIS)⇒ (WO)) Suppose that a non-empty subset S of N does not have a least element. Consider thestatement p(n), given by n 6∈ S. Then p(1) is true, otherwise 1 would be the least element of S. Supposenext that p(m) is true for every natural number m ≤ n, so that none of the numbers 1, 2, 3, . . . , n belongsto S. Then p(n + 1) must also be true, for otherwise n + 1 would be the least element of S. It nowfollows from (PIS) that S does not contain any element of N, contradicting the assumption that S is anon-empty subset of N.

Next, we complete the proof by showing that the Principle of induction (weak form) (PIW) is equivalentto the Principle of induction (strong form) (PIS).



((PIS) ⇒ (PIW)) Suppose that (PIW1) and (PIW2) both hold. Then clearly (PIS1) holds, since it isthe same as (PIW1). On the other hand, if p(m) is true for all m ≤ n, then p(n) is true in particular,so it follows from (PIW2) that p(n + 1) is true, and this gives (PIS2). It now follows from (PIS) thatp(n) is true for every n ∈ N.

((PIW)⇒ (PIS)) Suppose that (PIS1) and (PIS2) both hold for a statement p(.). Consider a statementq(.), where q(n) denotes the statement

p(m) is true for every m ≤ n.

Then the two conditions (PIS1) and (PIS2) for the statement p(.) imply respectively the two conditions(PIW1) and (PIW2) for the statement q(.). It follows from (PIW) that q(n) is true for every n ∈ N, andthis clearly implies that p(n) is true for every n ∈ N. ©

1.2. Completeness of the Real Numbers

The set Z of all integers is an extension of the set N of all natural numbers to include 0 and all numbersof the form −n, where n ∈ N. The set Q of all rational numbers is the set of all real numbers of theform pq−1, where p ∈ Z and q ∈ N. It is easy see that the Field axioms and Order axioms hold good ifthe set R is replaced by the set Q. We therefore need to find a property that distinguishes R from Q. Agood starting point is the following well known result.

THEOREM 1A. No rational number x ∈ Q satisfies x2 = 2.

Proof. Suppose that pq−1 has square 2, where p ∈ Z and q ∈ N. We may assume, without loss ofgenerality, that p and q have no common factors apart from ±1. Then p2 = 2q2 is even, so that p iseven. We can write p = 2r, where r ∈ Z. Then q2 = 2r2 is even, so that q is even, contradicting thatassumption that p and q have no common factors apart from ±1. ©

It follows that the real number we know as√

2 does not belong to the set Q. We say that the set Q isnot complete. Our idea is then to distinguish the set R from the set Q by completeness. In particular,we want to ensure that the set R contains numbers like

√2.

There are a number of ways to describe the completeness of the set R. We shall first of all introducecompleteness via the Axiom of bound.

Definition. A non-empty set S of real numbers is said to be bounded above if there exists a numberK ∈ R such that x ≤ K for every x ∈ S. The number K is called an upper bound of the set S.

Definition. A non-empty set T of real numbers is said to be bounded below if there exists a numberk ∈ R such that x ≥ k for every x ∈ T . The number k is called a lower bound of the set T .

AXIOM OF BOUND. Suppose that a non-empty set S of real numbers is bounded above. Then thereis a real number M ∈ R satisfying the following two conditions:

(S1) For every x ∈ S, the inequality x ≤M holds.(S2) For every ε > 0, there exists x ∈ S such that x > M − ε.

Remark. It is not difficult to prove that the number M above is unique. It is also easy to deduce thatif a non-empty set T of real numbers is bounded below, then there is a unique real number m ∈ Rsatisfying the following two conditions:

(I1) For every x ∈ T , the inequality x ≥ m holds.(I2) For every ε > 0, there exists x ∈ T such that x < m+ ε.



Definition. The real number M satisfying conditions (S1) and (S2) is called the supremum of thenon-empty set S, and denoted by M = supS. The real number m satisfying conditions (I1) and (I2) iscalled the infimum of the non-empty set S, and denoted by m = inf S.

Remark. Note that the most important point of the Axiom of bound is that the supremum M is a realnumber. Similarly, the infimum m is also a real number.

Let us now try to understand how numbers like√

2 fit into this setting. Recall that there is no rationalnumber which satisfies the equation x2 = 2. This means that the number that we know as

√2 is not a

rational number. We now want to show that it is a real number. Let

S = {x ∈ R : x2 < 2}.

Clearly the set S is non-empty, since 0 ∈ S. On the other hand, the set S is bounded above; for example,it is not difficult to show that if x ∈ S, then we must have x ≤ 2; for if x > 2, then we must have x2 > 4,so that x 6∈ S. Hence S is a non-empty set of real numbers and S is bounded above. It follows from theAxiom of bound that there is a real number M satisfying conditions (S1) and (S2). We shall show thatM2 = 2.

Suppose on the contrary that M2 6= 2. Then it follows from Axiom (O1) that M2 < 2 or M2 > 2.Let us investigate these two cases separately.

If M2 < 2, then we have

(M + ε)2 = M2 + 2Mε+ ε2 < 2 whenever ε < min{

1,2−M2

2M + 1

}.

This means that M + ε ∈ S, contradicting conndition (S1).

If M2 > 2, then we have

(M − ε)2 = M2 − 2Mε+ ε2 > 2 whenever ε <M2 − 2

2M.

This implies that any x > M − ε will not belong to S, contradicting condition (S2).

Note that M2 = 2 and M is a real number. It follows that what we know as√

2 is a real number.

Example 1.2.1. The set N is not bounded above but is bounded below with infimum 1.

Example 1.2.2. The set Z is not bounded above or below.

Example 1.2.3. The closed interval [√

2, 2] = {x ∈ R :√

2 ≤ x ≤ 2} is bounded above and below, withsupremum 2 and infimum

√2. Note that the supremum and infimum belong to the interval.

Example 1.2.4. The open interval (√

2, 2) = {x ∈ R :√

2 < x < 2} is bounded above and below, withsupremum 2 and infimum

√2. Note that the supremum and infimum do not belong to the interval.

Example 1.2.5. The set {x ∈ R : x = (−1)nn−1 for some n ∈ N} is bounded above and below, withsupremum 1/2 and infimum −1.

Example 1.2.6. The set {x ∈ Q : x2 < 2} is bounded above and below, with supremum√

2 and infimum−√2.

The argument concerning√

2 can be adapted to prove the following result.



THEOREM 1B. Suppose that a real number c ∈ R is positive. Then for every natural number q ∈ N,there exists a unique positive real number x ∈ R such that xq = c.

We denote by c1/q or q√c the unique positive real solution of the equation xq = c given by Theorem

1B. For every p ∈ Z and q ∈ N, we define cp/q = (c1/q)p. It can be shown that the definition of cm, wherem = p/q with p ∈ Z and q ∈ N, is independent of the choice of p and q. Furthermore, the Index laws aresatisfied: For every positive real number c ∈ R and rational numbers m,n ∈ Q, we have cmcn = cm+n

and (cm)n = cmn.

We next elaborate on Example 1.2.1, and prove formally that the set N is not bounded above. Thisis a consequence of the Axiom of bound.

THEOREM 1C. (ARCHIMEDEAN PROPERTY) For every real number x ∈ R, there exists a naturalnumber n ∈ N such that n > x.

Proof. Suppose that x ∈ R, and suppose on the contrary that n ≤ x for every n ∈ N. Then the set Nis bounded above by x, and so has a supremum M , say. In particular, we have

M ≥ 2, M ≥ 3, M ≥ 4, . . . ,

and soM − 1 ≥ 1, M − 1 ≥ 2, M − 1 ≥ 3, . . . .

Hence M − 1 is an upper bound for N, contradicting the hypothesis that M is the supremum of N. ©

We now establish the following important result central to the theory of mathematical analysis.

THEOREM 1D. The rational numbers and irrational numbers are dense in the set R. More precisely,between any two distinct real numbers, there exist a rational number and an irrational number.

Proof. Suppose that x, y ∈ R and x < y. We shall first show that there exists r ∈ Q such thatx < r < y. The idea is very simple. Heuristically, if we choose a natural number q large enough, thenthe interval (qx, qy) has length greater than 1 and must contain an integer p, so that qx < p < qy. Theformal argument is somewhat more complicated, but is based entirely on this idea.

Consider the special case when x > 0. By the Archimedean property, there exists q ∈ N such thatq > 1/(y−x), so that 1 < q(y−x). Consider the positive real number qx. By the Archimedean property,there exists n ∈ N such that n > qx. Using the Well-ordering principle, let p be the smallest such naturalnumber n. Then clearly p− 1 ≤ qx. To see this, note that if p = 1, then p− 1 = 0 < qx; if p 6= 1, thenp− 1 > qx would contradict the definition of p. It now follows that

qx < p = (p− 1) + 1 < qx+ q(y − x) = qy, so that x <p

q< y.

Suppose now that x ≤ 0. By the Archimedean property, there exists k ∈ N such that k > −x, so thatk + x > 0. There exists s ∈ Q such that x+ k < s < y + k, so that x < s− k < y. Clearly s− k ∈ Q.

To show that there exists z ∈ R \Q such that x < z < y, we first use our earlier argument twice, andconclude that there exist r1, r2 ∈ Q such that x < r1 < r2 < y. The number

z = r1 +1√2

(r2 − r1)

is clearly irrational and satisfies r1 < z < r2, and so x < z < y. ©.



1.3. The Complex Numbers

In this section, we briefly review some important properties of the complex numbers. It is easy to seethat the equation x2 +1 = 0 has no solution x ∈ R. In order to solve this equation, we have to introduceextra numbers into our number system.

Define the number i by i2 + 1 = 0. We then extend the field of all real numbers by adjoining thenumber i, which is then combined with the real numbers by the operations addition and multiplicationin accordance with the Field axioms in Section 1.1. The numbers a+ bi, where a, b ∈ R, of the extendedfield are then added and multiplied in accordance with the Field axioms, suitably extended, and therestriction i2 + 1 = 0. Note that the number a+ 0i, where a ∈ R, behaves like the real number a.

The set C = {z = x+ yi : x, y ∈ R} is called the set of all complex numbers. Note that in C, we losethe Order axioms and the Axiom of bound.

Suppose that z = x+yi, where x, y ∈ R. The real number x is called the real part of z, and denoted byx = Rez. The real number y is called the imaginary part of z, and denoted by y = Imz. Furthermore,we write

|z| =√x2 + y2

and call this the modulus of z.

Definition. A set S of complex numbers is said to be bounded if there exists a number K ∈ R suchthat |z| ≤ K for every z ∈ T .

THEOREM 1E. For every z, w ∈ C, we have(a) |zw| = |z||w|; and(b) |z + w| ≤ |z|+ |w|.

Proof. The first part is left as an exercise. To prove the Triangle inequality (b), note that the result istrivial if z + w = 0. Suppose now that z + w 6= 0. Then

|z|+ |w||z + w| =

|z||z + w| +

|w||z + w| =

∣∣∣∣ z

z + w

∣∣∣∣+∣∣∣∣ w

z + w

∣∣∣∣≥ Re

z

z + w+ Re

w

z + w= Re

(z

z + w+

w

z + w

)= Re1 = 1.

The result follows immediately. ©

Applying the Triangle inequality a finite number of times, we can show that for every z1, . . . , zk ∈ C,we have

|z1 + . . .+ zk| ≤ |z1|+ . . .+ |zk|.

We shall use this to establish the following result which shows that a polynomial is eventually dominatedby its term of highest order.

THEOREM 1F. Consider a polynomial P (z) = a0 + a1z + . . .+ anzn in the complex variable z ∈ C,

with coefficients a0, a1, . . . , an ∈ C and an 6= 0. For every z ∈ C satisfying

|z0| ≥ R0 =2(|a0|+ |a1|+ . . .+ |an|)

|an| ,

we have

12 |an||z|n ≤ |P (z)| ≤ 3

2 |an||z|n.



Proof. Note first of all that

|P (z)| ≤ |a0 + a1z + . . .+ an−1zn−1|+ |an||z|n

and

|an||z|n = |P (z)− (a0 + a1z + . . .+ an−1zn−1)| ≤ |P (z)|+ |a0 + a1z + . . .+ an−1z

n−1|.

It therefore remains to establish the inequality

|a0 + a1z + . . .+ an−1zn−1| ≤ 1

2 |an||z|n.

Clearly R0 > 1, so that if |z| ≥ R0, we have

|a0 + a1z + . . .+ an−1zn−1| ≤ |a0|+ |a1||z|+ . . .+ |an−1||z|n−1 ≤ (|a0|+ |a1|+ . . .+ |an−1|)|z|n−1

≤ (|a0|+ |a1|+ . . .+ |an−1|+ |an|)|z|n−1 = 12R0|an||z|n−1 ≤ 1

2 |an||z|n

as required. ©

1.4. Countability

In this brief account, we treat intuitively the distinction between finite and infinite sets. A set is finiteif it contains a finite number of elements. To treat infinite sets, our starting point is the set N of allnatural numbers, an example of an infinite set.

Definition. A set X is said to be countably infinite if there exists a bijective mapping from X to N. Aset X is said to be countable if it is finite or countably infinite.

Remark. Suppose that X is countably infinite. Then we can write

X = {x1, x2, x3, . . .}.

Here we understand that there is a bijective mapping φ : X → N where φ(xn) = n for every n ∈ N.

THEOREM 1G. A countable union of countable sets is countable.

Proof. Let I be a countable index set, where for each i ∈ I, the set Xi is countable. Either (a) I isfinite; or (b) I is countably infinite. We shall only consider (b), since (a) needs only minor modification.Since I is countably infinite, there exists a bijective mapping from I to N. We may therefore assume,without loss of generality, that I = N. For each n ∈ N, since Xn is countable, we may write

Xn = {an1, an2, an3, . . .},

with the convention that if Xn is finite, then the sequence an1, an2, an3, . . . is constant from some pointonwards. Hence we have a doubly infinite array

a11 a12 a13 . . .

a21 a22 a23 . . .

a31 a32 a33 . . .

......

.... . .



of elements of the set

X =⋃n∈N

Xn.

We now list these elements in the order indicated by

1–10 W W L Chen : Fundamentals of Analysis

of elements of the set

X =!n!N

Xn.

We now list these elements in the order indicated by

• • • • •

• • • •

• • • •

• • • •

but discarding duplicates. If X is infinite, the above clearly gives rise to a bijection from X to N. !

Example 1.4.1. The set Z is countable. Simply note that Z = N " {0} " {#1,#2,#3, . . .}.

Example 1.4.2. The set Q is countable. To see this, note that any x $ Q can be written in the formp/q, where p $ Z and q $ N. It is easy to see that for every n $ N, the set Qn = {p/n : p $ Z} iscountable. The result follows from Theorem 1G on observing that

Q =!n!N

Qn.

Suppose that two sets X1 and X2 are both countably infinite. Since both can be mapped to Nbijectively, it follows that each can be mapped to the other bijectively. In this case, we say that the twosets X1 and X2 have the same cardinality. Cardinality can be considered as a way of measuring size. Ifthere exists a one-to-one mapping from X1 to X2 and no one-to-one mapping from X2 to X1, then wesay that X2 has greater cardinality than X1. For example, N and Q have the same cardinality. We shallnow show that R has greater cardinality than Q.

To do so, we first need an intermediate result.

THEOREM 1H. Any subset of a countable set is countable.

Proof. Let X be a countable set. If X is finite, then the result is trivial. We therefore assume thatX is countably infinite, so that we can write

X = {x1, x2, x3, . . .}.

Let Y be a subset of X. If Y is finite, then the result is trivial. If Y is infinite, then we can write

Y = {xn1 , xn2 , xn3 , . . .},

where

n1 = min{n $ N : xn $ Y },

but discarding duplicates. If X is infinite, the above clearly gives rise to a bijection from X to N. ©

Example 1.4.1. The set Z is countable. Simply note that Z = N ∪ {0} ∪ {−1,−2,−3, . . .}.

Example 1.4.2. The set Q is countable. To see this, note that any x ∈ Q can be written in the formp/q, where p ∈ Z and q ∈ N. It is easy to see that for every n ∈ N, the set Qn = {p/n : p ∈ Z} iscountable. The result follows from Theorem 1G on observing that

Q =⋃n∈N

Qn.

Suppose that two sets X1 and X2 are both countably infinite. Since both can be mapped to Nbijectively, it follows that each can be mapped to the other bijectively. In this case, we say that the twosets X1 and X2 have the same cardinality. Cardinality can be considered as a way of measuring size. Ifthere exists a one-to-one mapping from X1 to X2 and no one-to-one mapping from X2 to X1, then wesay that X2 has greater cardinality than X1. For example, N and Q have the same cardinality. We shallnow show that R has greater cardinality than Q.

To do so, we first need an intermediate result.

THEOREM 1H. Any subset of a countable set is countable.

Proof. Let X be a countable set. If X is finite, then the result is trivial. We therefore assume that Xis countably infinite, so that we can write

X = {x1, x2, x3, . . .}.

Let Y be a subset of X. If Y is finite, then the result is trivial. If Y is infinite, then we can write

Y = {xn1 , xn2 , xn3 , . . .},

where

n1 = min{n ∈ N : xn ∈ Y },



and where, for every p ≥ 2,

np = min{n > np−1 : xn ∈ Y }.

The result follows. ©

THEOREM 1J. The set R is not countable.

Proof. In view of Theorem 1H, it suffices to show that the set [0, 1) is not countable. Suppose on thecontrary that [0, 1) is countable. Then we can write

[0, 1) = {x1, x2, x3, . . .}. (1)

For each n ∈ N, we express xn in decimal notation in the form

xn = .xn1xn2xn3 . . . ,

where for each k ∈ N, the digit xnk ∈ {0, 1, 2, . . . , 9}. Note that this expression may not be unique, butit does not matter, as we simply choose one. We now have

x1 = .x11x12x13 . . . ,

x2 = .x21x22x23 . . . ,

x3 = .x31x32x33 . . . ,

...

Let y = .y1y2y3 . . . , where for each n ∈ N, yn ∈ {0, 1, 2, . . . , 9} and yn ≡ xnn + 5 (mod 10). Then clearlyy 6= xn for any n ∈ N. But y ∈ [0, 1), contradicting (1). ©

Example 1.4.3. Note that the set R \ Q of all irrational numbers is not countable. It follows that inthe sense of cardinality, there are far more irrational numbers than rational numbers.

1.5. Cardinal Numbers

It is easy to show that there exists a bijective mapping from a finite set X1 to a finite set X2 if and onlyif the two sets X1 and X2 have the same number of elements. In this case, we say that the two setshave the same cardinality. It is then convenient to denote the cardinality of a finite set by the numberof elements that it contains, and take the non-negative integers to represent the finite cardinal numbers.

This may appear to be satisfactory. Strictly speaking, we need the following axiom which coversinfinite sets as well.

POSTULATE OF THE CARDINAL NUMBERS. For every set X, there exists an object |X|,called the cardinal number of X, which satisfies the following property: For any two sets X and Y , wehave |X| = |Y | if and only if there exists a bijective mapping f : X → Y .

Remarks. (1) Note that the cardinal number of an infinite set cannot be equal to the cardinal numberof a finite set, since there cannot be a bijective mapping from an infinite set to a finite set.

(2) We write ℵ0 = |N| and c = |R|.

(3) Note that |X| = ℵ0 for any countably infinite set X.

(4) In view of Theorem 1J, we have ℵ0 6= c.



Definition. Suppose that X and Y are sets.(1) We say that |X| ≤ |Y | if there exists an injective mapping f : X → Y .(2) We say that |X| < |Y | when |X| ≤ |Y | and |X| 6= |Y |.

Remarks. (1) Note that the definition is consistent with our observation at the beginning of this sectionand the usual meaning of the inequalities ≤ and < when applied to non-negative integers.

(2) Note that |X| < |Y | for every finite set X and infinite set Y .

The purpose of this section is to prove the following famous result. The special case when the sets Xand Y are finite is obvious.

THEOREM 1K. (CANTOR-BERNSTEIN-SCHRODER THEOREM) Suppose that X and Y are sets.Suppose further that |X| ≤ |Y | and |Y | ≤ |X|. Then |X| = |Y |.

Proof. Since |X| ≤ |Y | and |Y | ≤ |X|, there exist injective mappings f : X → Y and g : Y → X. Forevery x ∈ X, exactly one of the following holds:

• For every y ∈ Y , we have g(y) 6= x. In this case, we shall say that x has no predecessor.• There exists a unique y1 ∈ Y such that g(y1) = x. Here the uniqueness follows from the injective

property of the mapping g : Y → X. In this case, we shall say that y1 is the predecessor of x.

Similarly, for every y ∈ Y , exactly one of the following holds:

• For every x ∈ X, we have f(x) 6= y. In this case, we shall say that y has no predecessor.• There exists a unique x1 ∈ X such that f(x1) = y. Here the uniqueness follows from the injective

property of the mapping f : X → Y . In this case, we shall say that x1 is the predecessor of y.

Observe also that every x ∈ X is the predecessor of a unique element f(x) in Y , and that every y ∈ Y isthe predecessor of a unique element g(y) in X. It follows that for every element x ∈ X, we can constructa chain as follows:

. . .f−−−−→ y2

g−−−−→x1f−−−−→ y1

g−−−−→xf−−−−→ f(x)

g−−−−→ g(f(x))f−−−−→ . . .

Here y1 is the predecessor of x, x1 is the predecessor of y1, y2 is the predecessor of x1, and so on. Notethat the chain does not terminate on the right, but may terminate on the left at an element with nopredecessor. Similarly, for every element y ∈ Y , we can construct a chain as follows:

. . .g−−−−→x2

f−−−−→ y1g−−−−→x1

f−−−−→ yg−−−−→ g(y)

f−−−−→ f(g(y))g−−−−→ . . .

Here x1 is the predecessor of y, y1 is the predecessor of x1, x2 is the predecessor of y1, and so on.Again the chain does not terminate on the right, but may terminate on the left at an element with nopredecessor. It is easy to see that no element of X or Y can be in two distinct chains. We now define amapping h : X → Y as follows:

• For any element x ∈ X whose chain does not terminate on the left or terminates on the left withan element in X with no predecessor, we let h(x) = f(x).• For any element x ∈ X whose chain terminates on the left with an element of Y with no predecessor,

we let h(x) = y, where g(y) = x, so that y is the predecessor of x.

Note that the function h : X → Y defined in this way gives a one-to-one correspondence between theelements of X and the elements of Y in each chain, and so gives a one-to-one correspondence betweenthe elements of X and Y . ©



Problems for Chapter 1

1. Suppose that a, b ∈ R satisfy a > 0 and b < 0. Show that ab < 0.

2. Suppose that a, b ∈ R satisfy b < a < 0. Show that b−1 > a−1.

3. For each of the following sets A, determine whether supA and inf A exist, and find their values ifappropriate and determine also whether supA and inf A belong to the set A:

a) A = {n−1 : n ∈ N} b) A = {(|n|+ 1)−2 : n ∈ Z}c) A = {n+ n−1 : n ∈ N} d) A = {2−m − 3n : m,n ∈ N}e) A = {x ∈ R : x3 − 4x < 0} f) A = {1 + x2 : x ∈ R}

4. Suppose that A is a bounded set of real numbers, and that B is a non-empty subset of A. Explainwhy inf A ≤ inf B ≤ supB ≤ supA.

5. Suppose that a, b ∈ R satisfy a < b+ n−1 for every n ∈ N. Prove that a ≤ b.

6. a) Suppose that x ≤ a for every x ∈ A. Show that supA ≤ a.b) Show that the corresponding statement with ≤ replaced by < does not hold.

7. Suppose that A and B are non-empty sets of real numbers bounded above and below.a) Let A ∪B = {x : x ∈ A or x ∈ B}. Prove that

sup(A ∪B) = max{supA, supB} and inf(A ∪B) = min{inf A, inf B}.b) Discuss the case A ∩B = {x : x ∈ A and x ∈ B}.

8. Suppose that A and B are non-empty sets of real numbers bounded above and below.a) Let A+B = {a+ b : a ∈ A and b ∈ B}. Prove that

sup(A+B) = supA+ supB and inf(A+B) = inf A+ inf B.

b) Discuss the case A−B = {a− b : a ∈ A and b ∈ B}.

9. Suppose that A and B are non-empty sets of positive real numbers bounded above and below.a) Let AB = {ab : a ∈ A and b ∈ B}. Prove that

sup(AB) = (supA)(supB) and inf(AB) = (inf A)(inf B).

b) Discuss the case when the sets A and B can contain negative real numbers.

10. Suppose that A is a non-empty set of real numbers bounded above and below. For any real numberk ∈ R, consider the set kA = {ka : a ∈ A}. What can we say about sup(kA) and inf(kA)?

11. Prove that the cartesian product of two countable sets is countable.

12. A rational point in C is one with rational real and imaginary parts. Prove that the set of all rationalpoints in C is countable.

13. Prove that any isolated point set in C is countable.

14. a) Find a bijection from (0, 1) to (0,∞).b) Find a bijection from (−1, 1) to R.c) Suppose that A,B ∈ R with A < B. Find a bijection from (A,B) to (−1, 1).d) What is the cardinality of the interval (A,B) in part (c)?

15. A real algebraic number is any real solution of a polynomial equation with coefficients in Z. Provethat the set of all real algebraic numbers is countable.


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WWL Chen - Fundamentals of Analysis (Chapter 1)